Question: What is the slope of the line tangent to $f(x) = x^{2}-3x-5$ at $x = -2$ ?
The slope of the tangent line is $ \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$ $ = \lim_{h \to 0} \frac{((x+h)^{2}-3(x+h)-5) - (x^{2}-3x-5)}{h}$ $ = \lim_{h \to 0} \frac{(x^{2}+2x h+h^{2}-3(x+h)-5) - (x^{2}-3x-5)}{h}$ $ = \lim_{h \to 0} \frac{x^{2}+2(x h)+h^{2}-3x-3h-5-x^{2}+3x+5}{h}$ $ = \lim_{h \to 0} \frac{2(x h)+h^{2}-3h}{h}$ $ = \lim_{h \to 0} 2x+h-3$ $ = 2x-3$ $ = (2)(-2)-3$ $ = -7$